The sum of the terms of two $A . P$ are in the ratio $(3 n + 8); (7 n + 15)$ . Find the ratio of their $12 t h$ term.

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Solution

$\begin{matrix} \frac{S_{n}}{S_{n}^{1}} = \frac{3 n + 8}{7 n + 15} \to (i) S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d] + S_{n}^{1} = \frac{n}{2} [2 a_{1}^{1} + (n - 1) d^{1}] \frac{S_{n}}{S_{n}^{1}} = \frac{2 a_{1} + (n - 1) d}{2 a_{1}^{1} + (n - 1) d^{1}} = \frac{3 n + 8}{7 n + 15} s o, \frac{a_{1} + (\frac{n - 1}{2}) d}{a_{1}^{1} + (\frac{n - 1}{2}) d^{1}} = \frac{3 x + 8}{7 x + 15} \to (i i) s o, \frac{a_{m}}{a_{m}^{1}} = \frac{a_{1} + (m - 1) d}{a_{1}^{1} + (m - 1) d^{1}} s o, w e t a k e (\frac{n - 1}{2}) = m - 1 p u t t h e v a l u e o f n i n e q u a t i o n (i i) w e g e t n - 1 = 2 m - 2 n = 2 m - 1 \frac{a_{1} + (m - 1) d}{a_{1}^{1} + (m - 1) d^{1}} = \frac{3 (2 m - 1) + 8}{7 (2 m - 1) + 15} = \frac{6 m - 3 + 8}{14 m - 7 + 15} [\frac{a_{m}}{a_{m}^{1}} = \frac{6 m + 5}{14 m + 8}] p u t m = 12 \frac{a_{12}}{a_{12}^{1}} = \frac{6 (12) + 5}{14 (12) + 8} = \frac{77}{176} \end{matrix}$

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Similar questions

n

2

(3 n + 8) : (7 n + 15)

12^{t h}

3 n + 8 : 7 n + 15

The ratio between the sum of n terms of two A.Ps is 3n + 8 : 7 n + 15. Then the ratio between their $12^{t h}$ terms is

n

(3 n + 8) : (7 n + 15) .

(i) 12 t h

(i i) 15 t h

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